Download 1. Connectedness of a metric space A metric (topological) space X is

1. Connectedness of a metric space
A metric (topological) space X is disconnected if it is the union of two disjoint nonempty
open subsets. Otherwise, X is connected.
Theorem 1.1. Let X be a connected metric space and U is a subset of X. Assume that
(1) U is nonempty.
(2) U is closed.
(3) U is open.
Then U = X.
Proof. Suppose U 6= X. Then V = X \ U is nonempty. Since U is closed, V = X \ U is
open. Moreover, U ∩ V = ∅ and X = U ∪ V is a disjoint union of nonempty open subsets
of X. This implies that X is disconnected which leads to contradiction to the assumption
that X is connected.
Let X be a metric (topological) space and A be a subset of X. Then A is also a metric
(topological) space of A. Every open set in A is of the form U ∩ A for some open set U of
X.
We say that A is a (dis)connected subset of X if A is a (dis)connected metric (topological)
space.
Theorem 1.2. Let X and A be as above. Then A is disconnected if and only if there exist
open sets U, V in X so that
(1) U ∩ V ∩ A = ∅
(2) A ∩ U 6= ∅
(3) A ∩ V 6= ∅
(4) A ⊆ U ∩ V.
Proof. Suppose A is disconnected. Then A = U 0 ∪ V 0 , where U 0 , V 0 are disjoint open
nonempty subsets in A. Since A is a subspace of X, there are open sets U, V so that
U 0 = U ∩ A and V 0 = V ∩ A. Since A = U 0 ∪ V 0 , A ⊆ U ∪ V. Since U 0 and V 0 are nonempty,
U ∩ A and V ∩ A are nonempty. Moreover. U ∩ V ∩ A = U 0 ∩ V 0 is empty.
Conversely, suppose there are open subsets U, V obeying (1)-(4). Take U 0 = U ∩ A and
V 0 = V ∩ A. Then U 0 , V 0 are nonempty disjoint open subsets of A so that A = U 0 ∪ V 0 .
Hence A is disconnected.
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